| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola focus and directrix properties |
| Difficulty | Challenging +1.2 Part (a) is direct recall of parabola properties. Part (b) requires implicit differentiation and substitution but follows a standard method. Part (c) involves coordinate geometry to show diagonals intersect on the y-axis, requiring systematic algebraic manipulation across multiple steps, but the approach is methodical rather than requiring novel insight. This is moderately above average for A-level due to the extended multi-part nature and FP1 content. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(\frac{5}{2}, 0\right)\) | B1 | Deduces correct coordinates |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{5}{q}\) | B1 | Using or deriving \(\frac{dy}{dx} = \frac{5}{q}\) |
| At \(P\), \(x = \frac{q^2}{10}\) so tangent has equation \(y - q = \frac{5}{q}\left(x - \frac{q^2}{10}\right)\) | M1 | Finds equation of tangent using line formula with \(y_1 = q\), \(x_1 = \frac{q^2}{10}\) and \(m = \frac{2 \times \text{their } a}{q}\) |
| \(\Rightarrow qy - q^2 = 5x - \frac{q^2}{2} \Rightarrow 10x - 2qy + q^2 = 0\) * | A1\* | Completes correctly to given equation, no errors seen |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(B\) is \(\left(-\frac{5}{2}, q\right)\) | B1 | \(B\) is \(\left(-\frac{5}{2}, q\right)\) seen or used |
| Diagonal \(BF\) has equation \(\frac{y-0}{q-0} = \frac{x - \frac{5}{2}}{-\frac{5}{2} - \frac{5}{2}}\) or \(y = -\frac{q}{5}\left(x - \frac{5}{2}\right)\) | M1 | Correct method to find equation of diagonal \(BF\) using coordinates of \(F\) and \(B\) |
| \(10x - 2q\left(-\frac{q}{5}\left(x - \frac{5}{2}\right)\right) + q^2 = 0\) leading to \(\left\{y = \frac{25q + q^3}{50 + 2q^2}\right\}\) | dM1 | Uses printed answer in (b) and equation of diagonal \(BF\) to form equation involving \(x\) only, or solves simultaneously for \(y\) |
| \(\Rightarrow 10x + \frac{2q^2}{5}x - q^2 + q^2 = 0 \Rightarrow x\left(10 + \frac{2q^2}{5}\right) = 0\) | M1 | Correctly factors out \(x\) to achieve \(x(\ldots) = 0\) or uses expression for \(y\) to find expression for \(x\) |
| But \(10 + \frac{2q^2}{5} > 0\) so not zero, hence \(x = 0\), so intersection lies on \(y\)-axis | A1 | Conclusion given including reference to \(10 + \frac{2q^2}{5} \neq 0\) |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(x=0\) for \(BF\): \(y = -\frac{q}{5}\left(-\frac{5}{2}\right) = \ldots\) or for \(AP\): \(2qy = q^2 \Rightarrow y = \ldots\) | M1 | Attempts \(y\)-intercept for at least one diagonal |
| For \(BF\) \(y\)-intercept is \(\frac{q}{2}\) and for \(AP\) \(y\)-intercept is \(\frac{q}{2}\) | M1 | Finds \(y\)-intercept for both diagonals to compare |
| Since both diagonals always cross the \(y\)-axis at the same place, their intersection must always be on the \(y\)-axis | A1 | Both intercepts correct and suitable conclusion giving reference to both diagonals always crossing \(y\)-axis at same point |
| (9 marks total) |
# Question 4:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(\frac{5}{2}, 0\right)$ | **B1** | Deduces correct coordinates |
| | **(1)** | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = \frac{5}{q}$ | **B1** | Using or deriving $\frac{dy}{dx} = \frac{5}{q}$ |
| At $P$, $x = \frac{q^2}{10}$ so tangent has equation $y - q = \frac{5}{q}\left(x - \frac{q^2}{10}\right)$ | **M1** | Finds equation of tangent using line formula with $y_1 = q$, $x_1 = \frac{q^2}{10}$ and $m = \frac{2 \times \text{their } a}{q}$ |
| $\Rightarrow qy - q^2 = 5x - \frac{q^2}{2} \Rightarrow 10x - 2qy + q^2 = 0$ * | **A1\*** | Completes correctly to given equation, no errors seen |
| | **(3)** | |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $B$ is $\left(-\frac{5}{2}, q\right)$ | **B1** | $B$ is $\left(-\frac{5}{2}, q\right)$ seen or used |
| Diagonal $BF$ has equation $\frac{y-0}{q-0} = \frac{x - \frac{5}{2}}{-\frac{5}{2} - \frac{5}{2}}$ or $y = -\frac{q}{5}\left(x - \frac{5}{2}\right)$ | **M1** | Correct method to find equation of diagonal $BF$ using coordinates of $F$ and $B$ |
| $10x - 2q\left(-\frac{q}{5}\left(x - \frac{5}{2}\right)\right) + q^2 = 0$ leading to $\left\{y = \frac{25q + q^3}{50 + 2q^2}\right\}$ | **dM1** | Uses printed answer in (b) and equation of diagonal $BF$ to form equation involving $x$ only, or solves simultaneously for $y$ |
| $\Rightarrow 10x + \frac{2q^2}{5}x - q^2 + q^2 = 0 \Rightarrow x\left(10 + \frac{2q^2}{5}\right) = 0$ | **M1** | Correctly factors out $x$ to achieve $x(\ldots) = 0$ or uses expression for $y$ to find expression for $x$ |
| But $10 + \frac{2q^2}{5} > 0$ so not zero, hence $x = 0$, so intersection lies on $y$-axis | **A1** | Conclusion given including reference to $10 + \frac{2q^2}{5} \neq 0$ |
| | **(5)** | |
**Alternative for last three marks:**
| Answer | Mark | Guidance |
|--------|------|----------|
| When $x=0$ for $BF$: $y = -\frac{q}{5}\left(-\frac{5}{2}\right) = \ldots$ **or** for $AP$: $2qy = q^2 \Rightarrow y = \ldots$ | **M1** | Attempts $y$-intercept for at least one diagonal |
| For $BF$ $y$-intercept is $\frac{q}{2}$ **and** for $AP$ $y$-intercept is $\frac{q}{2}$ | **M1** | Finds $y$-intercept for both diagonals to compare |
| Since both diagonals always cross the $y$-axis at the same place, their intersection must always be on the $y$-axis | **A1** | Both intercepts correct and suitable conclusion giving reference to both diagonals always crossing $y$-axis at same point |
| | **(9 marks total)** | |
---\begin{enumerate}
\item The parabola $C$ has equation $y ^ { 2 } = 10 x$
\end{enumerate}
The point $F$ is the focus of $C$.\\
(a) Write down the coordinates of $F$.
The point $P$ on $C$ has $y$ coordinate $q$, where $q > 0$\\
(b) Show that an equation for the tangent to $C$ at $P$ is given by
$$10 x - 2 q y + q ^ { 2 } = 0$$
The tangent to $C$ at $P$ intersects the directrix of $C$ at the point $A$.\\
The point $B$ lies on the directrix such that $P B$ is parallel to the $x$-axis.\\
(c) Show that the point of intersection of the diagonals of quadrilateral $P B A F$ always lies on the $y$-axis.
\hfill \mbox{\textit{Edexcel FP1 AS 2022 Q4 [9]}}